An improper integral (Related to the Fourier transform)

[asmani]

How to show this?
\int_{-\infty}^{+\infty}e^{-i2\pi xs}ds=\delta(x)
This is a part of a problem of "Bracewell, R. The Fourier Transform and Its Applications, 3rd ed. New York: McGraw-Hill, pp. 100-101, 1999". This isn't a homework, I found it here (http://mathworld.wolfram.com/FourierTransform1.html). But I'm not sure even if this is true.

Thanks in advance.

[Mute]

Try showing that

\int_{-\infty}^{\infty} dx~\delta(x)f(x) = \int_{-\infty}^\infty dx \int_{-\infty}^\infty ds~e^-{2\pi i s x}f(x) = f(0).

[Dickfore]

\int_{-B}^{B}{e^{-i \, 2 \pi \, x \, s} \, ds} =\left.\frac{e^{-i \, 2 \pi \, x \, s}}{-i \, 2 \pi \, x}\right|_{B}^{B} = \frac{\sin{(\pi \, B \, x)}}{\pi \, x}


Next, consider the function:

\mathrm{sinc}(x) \equiv \frac{\sin{(\pi \, x)}}{\pi \, x}

and evaluate the definite integral (by contour integration)

\int_{-\infty}^{\infty}{\mathrm{sinc}(x) \, dx} = ?

Then, notice that the integral I evaluated is:

B \, \mathrm{sinc}(B \, x)

What happens if you take the limit B \rightarrow \infty?

[asmani]

Try showing that

\int_{-\infty}^{\infty} dx~\delta(x)f(x) = \int_{-\infty}^\infty dx \int_{-\infty}^\infty ds~e^-{2\pi i s x}f(x) = f(0).
I can show that if f is continuous at 0, then:
\int_{-\infty}^{+\infty}f(x)\delta(x)dx=f(0)
And if f has the Fourier transform, then:
\int_{-\infty}^{+\infty}\left [\int_{-\infty}^{+\infty}f(x)e^{-i2\pi xs}dx \right ]ds=f(0)
Does it help?

Thanks

[asmani]

\int_{-B}^{B}{e^{-i \, 2 \pi \, x \, s} \, ds} =\left.\frac{e^{-i \, 2 \pi \, x \, s}}{-i \, 2 \pi \, x}\right|_{B}^{B} = \frac{\sin{(\pi \, B \, x)}}{\pi \, x}


Next, consider the function:

\mathrm{sinc}(x) \equiv \frac{\sin{(\pi \, x)}}{\pi \, x}

and evaluate the definite integral (by contour integration)

\int_{-\infty}^{\infty}{\mathrm{sinc}(x) \, dx} = ?

Then, notice that the integral I evaluated is:

B \, \mathrm{sinc}(B \, x)

What happens if you take the limit B \rightarrow \infty?

I can see that if B \rightarrow \infty, then:

\int_{-\infty}^{\infty}{B\mathrm{sinc}(Bx) \, dx} = 1
But B\mathrm{sinc}(Bx) is not equal to 0 for every x≠0, while \delta (x) is.

Thanks

[asmani]

Actually the question is coming from here (http://www.physicsforums.com/showthread.php?p=3321756).